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Suppose we have an holomorphic function $$ f : \frac{\mathbb{C}}{\Lambda} \mapsto \frac{\mathbb{C}}{\Lambda} $$ where $\Lambda$ is a lattice. Is it always possible to find another function $\psi : \mathbb{C} \mapsto \mathbb{C}$ which extend $f$, i.e. such that $$ \forall x \in \mathbb{C}, \, \overline{\psi(x)} = f(\overline x) $$ or, at least, which extend $f$ on a neighbourhood of each point.

I met this issue when I was reading "rational points on elliptic curves" of Silverman & Tate. To give the context, we have an elliptic curve $C(\mathbb C)$ and an endomorphism of $C(\mathbb C)$, and because we know there exists an isomorphism $ \frac{\mathbb C}{\Lambda} \rightarrow C(\mathbb C)$ for a lattice $\Lambda$ (using Weierstrass $\wp$ function), we finally find an "holomorphic" (I don't really know what holomorphic on $\frac{\mathbb C}{\Lambda}$ mean) function $f$, and we use a function like $\psi$ and complex analysis to discover that $$ f : z \in \frac{\mathbb C}{\Lambda} \mapsto cz $$ with $c \in \mathbb C$. (Yes, it explain the name "complex multiplication")

You can find an article which explain this part of the book at : CMpaper

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Yes, given a function $f: \mathbb{C}/\Lambda \rightarrow \mathbb{C}/\Lambda$, define a new function $\tilde{f} \colon \mathbb{C} \rightarrow \mathbb{C}$ by setting $\tilde{f}(z) = z$ for $z$ in a fundamental parallelogram for $\Lambda$ and then extending it via periodicity to the plane. –  Barry Smith Apr 18 '11 at 15:05
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@Barry But is this doubly periodic? Because if $\tilde{f}$ is holomorphic and doubly periodic, then it is constant. If, on the other hand, $f$ has a pole somewhere on $\Lambda$, then $\tilde{f}$ will not be a map $\mathbb{C}\to \mathbb{C}$. –  Matt Apr 18 '11 at 16:23
    
If $p:X\to Y$ is a covering space, $Z$ is $1$-connected and $g:Z\to Y$ a continuous map then there is a lift $\tilde g:Z\to X$ (i.e. $p\circ \tilde g=f$). In your case $X\to Y=\mathbb{C}\to\mathbb{C}/\Lambda$, and $g:Z\to Y$ is the composition of $\mathbb{C}\to \mathbb{C}/\Lambda$ with $f$. (all functions are then easily seen to be holomorphic) –  user8268 Apr 18 '11 at 17:19
    
Thank you everybody, it took time to translate everything but I think I finally get it. Just like Matt said, Barry's construction can't work, it would contradict Liouville's theorem. Thank you user8268 for your helpful answer, I hadn't heard about covering space theory before! –  Valentin Apr 19 '11 at 15:15
    
Ah yes, I was thinking that a meromorphic function would answer his question, but reading it again, I see that he clearly wrote that he wants a function from $\mathbb{C}$ to $\mathbb{C}$. –  Barry Smith Apr 20 '11 at 15:11

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Answered by user8268:

If $p:X→Y$ is a covering space, $Z$ is $1$-connected and $g:Z→Y$ a continuous map then there is a lift $\tilde{g}:Z→X$ (i.e. $p∘\tilde{g}=f$). In your case $X→Y=\Bbb{C}→\Bbb{C}/Λ$, and $g:Z→Y$ is the composition of $\Bbb{C}→\Bbb{C}/Λ$ with $f$. (all functions are then easily seen to be holomorphic)

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